I am trying to prove why a natural number $n$ (which is not a multiple of $7$) when taken to the power of six ($n^6$) and divided by 7 always have the remainder of 1? I am not supposed to use "Fermats little theorem", but I am given the hint that the only numbers I have to account for are $1, 2, 3, 4, 5, 6$.

4 Answers
4

Every integer which is not divisible by $7$ is of the form $7k\pm 1$, $7k\pm 2$, or $7k\pm 3$, where $k$ is an integer. Imagine taking the $6$-th power of one of these numbers. The result will be a multiple of $7$, plus one of $1^6$, $2^6$, or $3^6$. It is easy to verify that in each of these cases, the result has remainder $1$ on division by $7$. For $1^6$ it is obvious. Since $2^6=64$, we have the result for $2^6$. For $3$ we note that $3^2$ has remainder $2$ on division by $7$, and $2^3$ has remainder $1$.

You can shrink the size of the numbers you need to deal with by noticing that
$$
n^6-1=(n^3-1)(n^3+1)=(n-1)(n+1)(n^2-n+1)(n^2+n+1) \, .
$$
So if any of those four factors are multiples of $7$, so too will be $n^6-1$.

Now,

$1-1=0 \times 7$

$2^2+2+1=1 \times 7$

$3^2-3+1=1 \times 7$

$4^2+4+1=3 \times 7$

$5^2-5+1=3 \times 7$

$6+1=1 \times 7$

so one of these factors is a multiple of $7$ for all $n \in \{1,2,3,4,5,6\}$.